Question: Solve for $r$, $ \dfrac{r + 5}{8r^3} = -\dfrac{10}{10r^3} - \dfrac{6}{4r^3} $
Explanation: First we need to find a common denominator for all the expressions. This means finding the least common multiple of $8r^3$ $10r^3$ and $4r^3$ The common denominator is $40r^3$ To get $40r^3$ in the denominator of the first term, multiply it by $\frac{5}{5}$ $ \dfrac{r + 5}{8r^3} \times \dfrac{5}{5} = \dfrac{5r + 25}{40r^3} $ To get $40r^3$ in the denominator of the second term, multiply it by $\frac{4}{4}$ $ -\dfrac{10}{10r^3} \times \dfrac{4}{4} = -\dfrac{40}{40r^3} $ To get $40r^3$ in the denominator of the third term, multiply it by $\frac{10}{10}$ $ -\dfrac{6}{4r^3} \times \dfrac{10}{10} = -\dfrac{60}{40r^3} $ This give us: $ \dfrac{5r + 25}{40r^3} = -\dfrac{40}{40r^3} - \dfrac{60}{40r^3} $ If we multiply both sides of the equation by $40r^3$ , we get: $ 5r + 25 = -40 - 60$ $ 5r + 25 = -100$ $ 5r = -125 $ $ r = -25$